Question

A stone thrown into still water, creates a circular wave pattern moving radially outwards. If r is the distance measured from the centre of the pattern. The amplitude of the wave varies as
A) ${r^{ - \dfrac{1}{2}}}$
B) ${r^{ - 1}}$
C) ${r^{ - 2}}$
D) ${r^{ - \dfrac{3}{2}}}$

Answer 1

When a circular wave pattern is created, there are certain relationships among the quantities. Here the distance from the centre (r) is given and variation of amplitude (A) is to be found but there is no direct relationship between the two so we can use intensity as a mediocre who has a relationship with both A and r.

Complete step by step answer:
When a stone is thrown into still water, a circular wave pattern is generated:

Intensity (I) : It is defined as the power of a wave per unit area. It is inversely proportional to the distance measured from the centre (r) and directly proportional to the square of the amplitude. Mathematically:
$I \propto \dfrac{1}{r}$ and $I \propto {A^2}$
Amplitude (A) : It is the maximum height that has been covered by a wave and is directly proportional to the square root of intensity.
$A \propto \sqrt I$
The waves form are circular and the relationship between intensity – radius and intensity – amplitude is given as:
$I \propto \dfrac{1}{r}$ or
$I \propto {r^{ - 1}}$ _____ (1)
And
$A \propto \sqrt I$
Substituting the value of I from (1), we get:
$A \propto \sqrt {\dfrac{1}{r}}$ or
$A \propto \dfrac{1}{{{r^{\dfrac{1}{2}}}}}$ or
$A \propto {r^{ - \dfrac{1}{2}}}$
This shows that the amplitude of the circular waves formed is directly proportional to and hence varies as ${r^{ - \dfrac{1}{2}}}$. Therefore, the correct option is option A).

Note: Variation of two quantities with respect to each other can be given as:
$a \propto b$ : Shows that ‘a’ is directly proportional to ‘b’ which means with the increase in ‘a’, quantity ‘b’ also increases and vice-versa.
$a \propto \dfrac{1}{b}$ : Shows that ‘a’ is inversely proportional to ‘b’ which means with the increase in ‘a’, quantity ‘b’ decreases and vice - versa.